Question: Consider the parametric curve: $\begin{aligned} x&=\ln(t+5) \\\\ y&=-6t^{5} \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=-2$ to $t=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{-2}^{3} \sqrt{\dfrac{1}{t+5}-30t^4}\,dt$ (Choice B) B $\int_{-2}^{3} \sqrt{\ln(t+5)-6t^{5}}\,dt$ (Choice C) C $\int_{-2}^{3} \sqrt{\dfrac{1}{(t+5)^2}+900t^{8}}\,dt$ (Choice D) D $\int_{-2}^{3} \sqrt{(\ln(t+5))^2+36t^{10}}\,dt$
Solution: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[\ln(t+5)\right] \\\\ &=\dfrac{1}{t+5} \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[-6t^{5}\right] \\\\ &=-30t^4 \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{-2}^{3} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{-2}^{3} \sqrt{\left(\dfrac{1}{t+5}\right)^2+\left(-30t^4\right)^2}\,dt \\\\ &=\int_{-2}^{3} \sqrt{\dfrac{1}{(t+5)^2}+900t^{8}}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=-2$ to $t=3$ : $\int_{-2}^{3} \sqrt{\dfrac{1}{(t+5)^2}+900t^{8}}\,dt$